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[Sorry if the level here is wrong, I asked this on math.SE, but even with a bounty, it got no attention.]

I am currently reading Hatcher's 3-Manifolds notes, the part proving Alexander's theorem, which is a specific case of the generalized Schoenflies theorem:

Every smoothly embedded $S^2\subset \mathbb{R}^3$ bounds a smooth 3-ball.

The proof seems to rely on intuition for these low-dimensional arguments, which I find disconcerting, because I have not yet developed that intuition, so I am trying to give actual formal proofs for the statements in Hatcher's proof.

The proof begins with a generic smoothly embedded closed surface $S\subset\mathbb{R}^3$. I have been able to prove that I can isotope $S$ so that projection on the last coordinate $\pi:\mathbb{R}^3\rightarrow\mathbb{R}$ is a Morse function on $S$. Hatcher then argues that if $t$ is a regular value for $\pi$, then $\pi^{-1}(t)\cap S$ is a finite collection of circles.

The proof continues by taking an innermost circle $C\subset \pi^{-1}(t)\cap S$, which by 2-dimensional Schoenflies bounds a disk $D$, and $D\cap S=\partial D=C$. Hatcher then uses surgery to cut away a neighborhood of $C$ in $S$, and cap the cuts with two disks.

This last part is what I want to formalize. It seems we are finding a small-enough tubular neighborhood $C\times(-\epsilon,\epsilon)\subset S$, and then removing that, leaving $S_-=C\times\{-\epsilon\}$ and $S_+=C\times\{\epsilon\}$. Again by 2-dimensional Schoenflies, these bound disks $D_-$ and $D_+$. What I don't get is: why are $S_-$ and $S_+$ still innermost? Or, put differently, why is $D_-\cap S=S-$ (and similarly for $S_+$)?

Intuitively, this seems obvious, and it seems like some sort of "continuity" argument would work, but I cannot figure out how to make this formal. I tried proving that in fact all the disks, "stacked" together for the tubular neighborhood, gave a smooth $D\times [-\epsilon,\epsilon]$, but again I find it hard to make topological arguments when one step of the construction is "apply Schoenflies to get a disk". In particular, I can't prove the projection of this "solid neighborhood" to $D$ is continuous.

Does anyone know how to formalize this? Or, even better, a reference where this type of surgery is discussed? I checked a few places, but only found surgery on a single manifold, not the type discussed here, where we're surgering an embedded submanifold in some ambient manifold.

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If $t$ is a regular value, then it is a property of Morse functions that there is some small open neighborhood $U$ of $t$ in $\mathbb{R}$ such that $u$ is also a regular value for all $u\in U$. In particular, we can take $U=(t-\delta,t+\delta)$ for some $\delta>0$. But then $\pi^{-1}(U)\cap S\cong(\pi^{-1}(t)\cap S)\times U$, ie. the surface is a product between any two successive critical levels. Now simply choose $\epsilon<\delta$. Then $S_-=(C\times U)\cap\pi^{-1}(t-\epsilon)$ and $S_+=(C\times U)\cap\pi^{-1}(t+\epsilon)$ will be innermost since $C$ is innermost.

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  • $\begingroup$ Sorry, I don't follow the final sentence at all. Why would there be a critical value? $\endgroup$ – Hempelicious Feb 3 at 3:54
  • $\begingroup$ I deleted that sentence. Is it clear now? $\endgroup$ – Josh Howie Feb 3 at 4:55
  • $\begingroup$ I think it's intuitively clear, and does help me understand things better, but I'm trying to avoid intuition. In particular, your diffeomorphism with the product does not take into account the ambient slices, so I don't see why we can say anything about a circle being innermost or not. Here's an example of what I'm worried about: if all we know at each slice is there are two circles, there's nothing a priori preventing one moving "inside" another. Intuitively, that can't happen because the enclosed disks would change how they intersect. It's this last part I can't formalize. $\endgroup$ – Hempelicious Feb 3 at 19:35
  • $\begingroup$ That cannot happen because there is also a product $D\times U$. $\endgroup$ – Josh Howie Feb 3 at 22:21
  • $\begingroup$ I think that's the Crux of the issue: how do we know that? How do we know the disks we get from Jordan curve theorem can be aligned as a product via the map you've provided? $\endgroup$ – Hempelicious Feb 3 at 23:52

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